I need DISPLACEMENT please.<br><br>I have Average speed, i need Displacement.

9) What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

Hunter-Best [27]

The weight of the astronaut is given by
$W=mg$
where m=59.1 kg is his mass and $g=9.81~m/s^2$ is the gravitational acceleration on Earth.

To solve the problem, we must find the value of g on the new planet. g is given by
$g= \frac{GM}{r^2}$
where G is the gravitational constant, M the mass of the planet and r its radius.
The mass of the planet can be written as
$M=dV$
where d is the density and V the volume.
We can assume that the planet is a sphere, therefore the volume is proportional to $r^3$ :
$V= \frac{4}{3}\pi r^3$
and we can write the mass as
$M= \frac{4}{3} \pi d r^3$
and then, g becomes
$g= \frac{GM}{r^2}= \frac{4}{3} \frac{G \pi d r^3}{r^2}= \frac{4}{3} G \pi d r$
So, in the end g is proportional to the radius of the planet, r (because the density of the new planet d is the same as the Earth's one. If the radius of the new planet is twice the Earth's radius, g will be twice the value of g on Earth:
$g_{new}=2g=2\cdot9.81~m/s^2=19.62~m/s^2$
And since the mass of the astronaut is always the same, the weight on the new planet will be twice the weight on Earth:
$W_{new}=mg_{new}=2mg=1159~N$

5 0

3 years ago

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

8 0

2 years ago

Motivation is best described as the _____. ability to be successful with your exercise way you see your exercise desire to conti

Olin [163]

Answer:

The desire to continue with your exercise would be the correct answer! :D

6 0

3 years ago

Read 2 more answers

What is the SI unit for momentum?

serg [7]

It’s a vector quantity, which means it possesses both magnitude and direction. So the SI unit would be B)kg•m/s

4 0

3 years ago

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An object traveling at a constant speed but with a changing direction is accelerating.

prohojiy [21]

Strange as it may seem, that's true. (choice 'a'.)

"Acceleration" doesn't mean "speeding up". It means ANY change in
the speed or direction of motion. So a car with the brakes applied
and slowing down, and a point on the rim of a bicycle wheel that's
turning at a constant rate, are both accelerating.

6 0

3 years ago

Read 2 more answers

I need DISPLACEMENT please.I have Average speed, i need Displacement.